The velocity of the aircraft relative to the ground is 240 km/h North
Explanation:
We can solve this problem by using vector addition. In fact, the velocity of the aircraft relative to the ground is the (vector) sum between the velocity of the aircraft relative to the air and the velocity of the air relative to the ground.
Mathematically:

where
v' is the velocity of the aircraft relative to the ground
v is the velocity of the aircraft relative to the air
is the velocity of the air relative to the ground.
Taking north as positive direction, we have:
v = +320 km/h
(since the air is moving from North)
Therefore, we find
(north)
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Answer:
3.5 cm
Explanation:
mass, m = 50 kg
diameter = 1 mm
radius, r = half of diameter = 0.5 mm = 0.5 x 10^-3 m
L = 11.2 m
Y = 2 x 10^11 Pa
Area of crossection of wire = π r² = 3.14 x 0.5 x 10^-3 x 0.5 x 10^-3
= 7.85 x 10^-7 m^2
Let the wire is stretch by ΔL.
The formula for Young's modulus is given by


ΔL = 0.035 m = 3.5 cm
Thus, the length of the wire stretch by 3.5 cm.
So the equation for angular velocity is
Omega = 2(3.14)/T
Where T is the total period in which the cylinder completes one revolution.
In order to find T, the tangential velocity is
V = 2(3.14)r/T
When calculated, I got V = 3.14
When you enter that into the angular velocity equation, you should get 2m/s
Answer:
There is an inward force acting on the can
Explanation:
This inward force is known as Centripetal force and it is responsible for making the can whirl on the end of a string in circle and it is also directed towards the center around which the can is moving.
Note that
1 km/h = (1000 m)/(3600 s) = 0.27778 m/s
Initial velocity, v₁ = 25 km/h = 6.9444 m/s
Final velocity, v₂ = 65 km/h = 18.0556 m/s
Time interval, dt = 6 s.
Calculate average acceleration.
a = (v₂ - v₁)/dt
= (18.0556 - 6.9444 m/s)/(6 s)
= 1.852 m/s²
Answer:
The average acceleration is 1.85 m/s² (nearest hundredth)